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Zigmanuir [339]
3 years ago
5

3 examples of density synonyms

Chemistry
1 answer:
DENIUS [597]3 years ago
3 0

frequency.

quantity.

thickness.

body.

closeness.

concretion.

heaviness.

solidity.

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Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas
IrinaK [193]
<span>convert grams to moles then use the equatino to do moles of Na to moles of H2O then convert moles of H2O to molecules by using avogadros number (6.022e23)

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7 0
2 years ago
To produce large tomatoes that are resistant to cracking and splitting some seeds companies use the pollen from one variety of t
rodikova [14]

Answer:

cross pollination

Explanation:

is when two types of plants are mixed to create a better or hardier plant

4 0
3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
If the half-life of Carbon-14 is 5700 years, how many years would it take a sample to decay from 1 gram to 31.3 mg
andrew-mc [135]

Answer:

28500 years

Explanation:

Applying,

A = A'(2^{x/y})............... Equation 1

Where A = Original mass of Carbon-14, A' = Final mass of carbon-14 after decaying, x = total time, y = half-life.

From the question,

Given: A = 1 g, A' = 31.3 mg = 0.0313 g, y = 5700 years.

Substitute these values into equation 1

1 = 0.0313(2^{x/5700})

2^{x/5700} = 1/0.0313

2^{x/5700}  = 31.95

2^{x/5700} ≈ 32

2^{x/5700} ≈ 2⁵

Equating the base and solve for x

x/5700 ≈ 5

x ≈ 5×5700

x ≈ 28500 years

3 0
2 years ago
Which metal will react spontaneously with Cu2+ (aq) at 25°C?
Gwar [14]

Answer:

Mg  

Explanation:

The standard reduction potentials are

                                              <u>E°/V </u>

Au³⁺(aq ) + 3e⁻  ⟶  Au(s);     1.42

Hg²⁺(aq)  + 2e⁻  ⟶  Hg(l);     0.85

Ag⁺(aq)    +   e⁻  ⟶  Ag(s);    0.80

Cu²⁺(aq)   + 2e⁻ ⟶  Cu(s);   0.34

Mg2+(aq) + 2e- ⟶  Mg(s);   -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

 

5 0
2 years ago
Read 2 more answers
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